INTRODUCTION to SPECTRAL SEQUENCES 1. Introduction in Homological Algebra and Algebraic Topology, a Spectral Sequence Is a Means

INTRODUCTION TO SPECTRAL SEQUENCES

HUAN VO

1. Introduction In homological algebra and algebraic topology, a spectral sequence is a means of computing homology groups by taking successive approximations. Roughly speaking, a spectral sequence is a r r sequence (E , d )r≥1 of bicomplexes, which is can think of as a “book” with infinitely many “pages”. Every time we go to the next page, we get a bit closer to the homology of the complex we want to compute. Spectral sequences were invented in the 1940s, independently, by J. Leray and R. C. Lyndon. Leray was a German prisoner of war from 1940 through 1945, during World War II. This notes aims to give a brief introduction to spectral sequences from a purely homological algebra point of view. Specifically, we only describe the set-up of spectral sequences, without going into their many exciting applications in topology and geometry. For this notes, we closely follow the treatment found in [1]. We only assume very basic homological algebra facts. In particular, we assume the readers are comfortable with the notions of complexes, homology, exact sequence in homologies. Some technical proofs which are not very essential to understanding the flow of ideas are referenced in [1]. The organization of the notes goes as follows. In Section 2, we introduce our objects of study: bicomplexes and their total complexes. Our main goal is to compute the homology of the total complex of a complex. In Section 3, we describe the essential concept of a filtration of a complex and how to go from a filtration of a complex to an exact couple, which then gives rise to spectral sequence, by a recursive process. In Section 4, we discuss spectral sequences and show that a spectral sequence arising from a filtration of a complex “converges” to the homology of the complex. Here by convergence, we mean only up to extension, i.e. there exists a short exact sequence p−1 p ∞ 0 −→ Φ Hn −→ Φ Hn −→ Ep,q −→ 0. In Section 5, we apply our general discussion to the case of the total complex of a bicomplex and describe a concrete way to compute the first two pages of the spectral sequence using iterated homologies. Finally, in Section 6, we illustrate how to use the machinery we have developed to show that the definition of the Tor functor is independent of the variable resolved. In this notes R will denote a commutative ring with 1. It follows that left R-modules are the same as right R-modules, and we just call them R-modules.

2. Bicomplexes

Definition 2.1. A graded module is an indexed family M = (Mp)p∈Z of R-modules. Definition 2.2. Let M, N be graded modules and let a ∈ Z .A graded map of degree a, denoted by f : M → N, is a family of R-module homomorphisms f = (fp : Mp → Np+a)p∈Z . The degree of f is a, and we denote it by deg(f) = a. Let M and M 0 be graded modules, then M 0 is a submodule of M, denoted M 0 ⊆ M, if each 0 0 Mp is a submodule of Mp for all p ∈ Z . If M is a submodule of M, then we can form the 0 0 quotient M/M = (Mp/Mp)p∈Z . If f : M → N is a graded map of degree a, then the kernel 1 INTRODUCTION TO SPECTRAL SEQUENCES 2 of f is the graded module ker f = (ker fp)p∈Z ⊆ M and the image of f is the graded module im f = (im fp−a)p∈Z ⊆ N. A sequence of graded maps f g A −→ B −→ C is exact if im f = ker g, i.e. if im fp−a = ker gp for all p ∈ Z . Definition 2.3. A complex (or chain complex) is an ordered pair (C, d), where C is a graded module and d is a graded map of degree −1 that satisfies dd = 0. The map d is called the differential.

dn+2 dn+1 dn dn−1 C : ··· −−−→ Cn+1 −−−→ Cn −→ Cn−1 −−−→· · · .

The condition dd = 0 implies that im dn+1 ⊆ ker dn for all n ∈ Z . The nth homology module of (C, d) is deﬁned as

Hn(C) = ker dn/im dn+1.

Deﬁnition 2.4. A bigraded module is a doubly indexed family M = (Mp,q)(p,q)∈Z ×Z of R- modules.

Definition 2.5. Let M, N be bigraded modules and let (a, b) ∈ Z × Z .A bigraded map of degree (a, b), denoted by f : M → N, is a family of R-module homomorphisms f = (fp,q : Mp,q → Np+a,q+b)(p,q)∈Z ×Z . The bidegree of f is (a, b), and we denote it by deg(f) = (a, b). Let M and M 0 be bigraded modules, then M 0 is a submodule of M, denoted M 0 ⊆ M, if each 0 0 Mp,q is a submodule of Mp,q for all (p, q) ∈ Z × Z . If M is a submodule of M, then we can form 0 0 the quotient M/M = (Mp,q/Mp,q)(p,q)∈Z ×Z . If f : M → N is a bigraded map of bidegree (a, b), then the kernel of f is the bigraded module ker f = (ker fp,q)(p,q)∈Z ×Z ⊆ M and the image of f is the bigraded module im f = (im fp−a,q−b)(p,q)∈Z ×Z ⊆ N. A sequence of bigraded maps f g A −→ B −→ C is exact if im f = ker g, i.e. if im fp−a,q−b = ker gp,q for all (p, q) ∈ Z × Z . Definition 2.6. A differential bigraded module is an ordered pair (M, d), where M is a bigraded module and d: M → M is a bigraded map with dd = 0. The map d is called the differential. If (M, d) is a bigraded module, where d has bidegree (a, b), then its homology H(M, d) is the bigraded module whose (p, q) term is

ker dp,q H(M, d)p,q = . im dp−a,q−b Definition 2.7. A bicomplex (or double complex) is an ordered triple (M, d0, d00), where M = 0 00 (Mp,q) is a bigraded module, d , d : M → M are bigraded maps of bidegree (−1, 0) and (0, −1), respectively that satisfy (i) d0d0 = 0, (ii) d00d00 = 0, 0 00 00 0 (iii) dp,q−1dp,q + dp−1,qdp,q = 0 0 00 Thus we can visualize the bicomplex (M, d , d ) as follows. Put Mp,q at the lattice point (p, q) 0 00 of Z × Z . The map d points to the left and the map d points down. The first condition says that each row forms a complex with differentials d0 and the second condition says that each column forms a complex with differentials d00. The third condition says that each square of the lattice anticommutes. INTRODUCTION TO SPECTRAL SEQUENCES 3

Deﬁnition 2.8. If (M, d0, d00) is a bicomplex, then its total complex, denoted by Tot(M), is the complex with nth term M Tot(M)n = Mp,q p+q=n P 0 00 and with diﬀerentials Dn : Tot(M)n → Tot(M)n−1 given by Dn = p+q=n(dp,q + dp,q).

0 dp−1,q+1 Mp−2,q+1 Mp−1,q+1 00 dp−1,q+1 0 dp,q Mp−1,q Mp,q 00 dp,q

Mp,q−1

Lemma 2.9. If (M, d0, d00) is a bicomplex, then (Tot(M),D) is a complex.

Proof. We need to check that indeed Dn : Tot(M)n → Tot(M)n−1 and Dn−1Dn = 0. This is where condition (iii) in deﬁnition 2.7 comes into play. The readers are encouraged to draw the picture and see that directly. 3. Filtrations and Exact Couples

Definition 3.1. A filtration of an R-module M is a family (Mp)p∈Z of submodules of M such that · · · ⊆ Mp−1 ⊆ Mp ⊆ Mp+1 ⊆ · · · . For each p ∈ Z , the module Mp/Mp−1 is called a factor module of the filtration. p Definition 3.2. A filtration of a complex (C, d) is a sequence of graded submodules (F C)p∈Z of C such that · · · ⊆ F p−1C ⊆ F pC ⊆ F p+1C ⊆ · · · that satisfies p p d:(F C)n → (F C)n−1 p p for all n, p. To simplify notation, we denote (F C)n by Fn . Note that the following diagram commutes......

p−1 p p+1 ··· Fn+1 Fn+1 Fn+1 ···

p−1 p p+1 ··· Fn Fn Fn ···

p−1 p p+1 ··· Fn−1 Fn−1 Fn−1 ··· ......

p The horizontal maps are inclusions and the vertical maps are restrictions of d to Fn . The nth row form a ﬁltration of Cn. INTRODUCTION TO SPECTRAL SEQUENCES 4

Definition 3.3. Let (M, d0, d00) be a bicomplex. The first filtration (or vertical filtration) of Tot(M) is given by I p M ( F Tot(M))n = Mi,n−i i≤p

= · · · ⊕ Mp−2,q+2 ⊕ Mp−1,q+1 ⊕ Mp,q. Here we use the convention that n = p + q. Thus we have I p−1 I p I p+1 · · · ⊆ ( F Tot(M))n ⊆ ( F Tot(M))n ⊆ ( F Tot(M))n ⊆ · · · I p a filtration of Tot(M)n. Pictorially, ( F Tot(M))n is the direct sum of the terms on the line x + y = n which lie to the left of the vertical line x = p. Definition 3.4. Let (M, d0, d00) is a bicomplex. The second filtration of Tot(M) (or horizontal filtration) is given by II p M ( F Tot(M))n = Mn−j,j j≤p

= · · · ⊕ Mq−1,p−2 ⊕ Mq+1,p−1 ⊕ Mq,p, where we again use the convention p + q = n. We have II p−1 II p II p+1 · · · ⊆ ( F Tot(M))n ⊆ ( F Tot(M))n ⊆ ( F Tot(M))n ⊆ · · · II p a filtration of Tot(M)n. Pictorially, ( F Tot(M))n is the direct sum of the terms on the line x + y = n which lie below the horizontal line y = p. Exercise 3.5. Check that the above definitions indeed give us filtrations of Tot(M), i.e. check p p that d:(F Tot(M))n → (F Tot(M))n−1. To see the relationship between the first filtration and the second filtration, we introduce the concept of the transpose of a bicomplex. Definition 3.6. Let (M, d0, d00) be a bicomplex. The transpose of (M, d, d00) is the bicomplex t 0 00 t 0 00 00 0 (M , δ , δ ), where Mp,q = Mq,p, δp,q = dq,p and δp,q = dq,p. Lemma 3.7. Let (M, d0, d00) be a bicomplex. Then (i) Tot(M) = Tot(M t), (ii) the second filtration of Tot(M) is equal to the first filtration of Tot(M t), i.e. II p I p t F Tot(M)n = F Tot(M )n. Proof. The proof consists of manipulating the indices. We have t M t M Tot(M )n = Mp,q = Mq,p = Tot(M)n. p+q=n p+q=n Also, II p M M t I p t F Tot(M)n = Mn−j,j = Mj,n−j = F Tot(M )n, j≤p j≤p as required. Definition 3.8. An exact couple is a 5-tuple (D, E, α, β, γ), where D and E are bigraded modules, α, β, γ are bigraded maps with bidegrees (a, a0), (b, b0) and (c, c0), respectively, and there is exactness at each vertex: ker α = im γ, ker β = im α, and ker γ = im β. Thus an exact couple is equivalent to a collection of exact sequences: for each (p, q) ∈ Z × Z we have an exact sequence:

· · · −→ Ap,q −→ Bp+a,q+a0 −→ Cp+a+b,q+a0+b0 −→ Ap+a+b+c,q+a0+b0+c0 −→ · · · INTRODUCTION TO SPECTRAL SEQUENCES 5

α D D

γ β E

α (1, −1) D D

γ (−1, 0) β (0, 0) E

p Proposition 3.9. Every ﬁltration (F C)p∈Z of a complex (C, d) determines an exact couple whose bigraded maps have the displayed bidegrees. p p Proof. To simplify notation, we write F C as F . For each p ∈ Z , we have the following short exact sequence of complexes: 0 −→ F p−1 −→ F p −→ F p/F p−1 −→ 0, or in expanded form......

p−1 p p p−1 0 Fn+1 Fn+1 Fn+1/Fn+1 0

p−1 p p p−1 0 Fn Fn Fn /Fn 0

p−1 p p p−1 0 Fn−1 Fn−1 Fn−1/Fn−1 0 ......

This induces a long exact sequence in homologies (see [1, Theorem 6.10]).

p−1 p p p−1 ··· Hn+1(F ) Hn+1(F ) Hn+1(F /F ) δ

p−1 p p p−1 Hn(F ) Hn(F ) Hn(F /F ) δ

p−1 p p p−1 Hn−1(F ) Hn−1(F ) Hn−1(F /F ) ···

Now for each (p, q) ∈ Z × Z , put p p p−1 Dp,q = Hp+q(F ),Ep,q = Hp+q(F /F ). We then obtain a long exact sequence.

· · · −→ Dp−1,q+2 −→ Dp,q+1 −→Ep,q+1 −→ Dp−1,q+1 −→ Dp,q −→

−→ Ep,q −→ Dp−1,q −→ Dp,q−1 −→ · · · , INTRODUCTION TO SPECTRAL SEQUENCES 6 which is equivalent to the exact couple

α (1, −1) D D

γ (−1, 0) β (0, 0) E

Proposition 3.10. If (D, E, α, β, γ) is an exact couple, where α, β, γ have bidegrees (a, a0), (b, b0), (c, c0), respectively, then d1 = βγ is a diﬀerential d1 : E → E, and there is an exact couple (D2,E2, α2, β2, γ2), called the derived couple, with E2 = H(E, d1), and whose bigraded maps have the displayed bidegrees.

α2 (a, a0) D2 D2

γ2 (c, c0) β2 (b − a, b0 − a0) E2

Proof. We refer the readers to [1, Proposition 10.9] for a proof. For our purpose, we just need to know the definitions of the terms in the derived couple. We have E2 = H(E, d1) and D2 = im α ⊆ D. Now the bigraded map α2 : D2 → D2 is the restriction of α to D2; the bigraded map β2 : D2 → E2 = H(E, d1) is given by 2 −1 2 β (y) = [βα (y)] for y ∈ Dp,q; and the bigraded map γ2 : E2 → D2 is given by γ2([z]) = γz 2 for [z] ∈ Ep,q. The above procedure can be iterated, i.e. given (D2,E2, α2, β2, γ2), we can obtain an exact couple (D3,E3, α3, β3, γ3), with E3 = H(E2, d2). More generally, we define the (r + 1)st derived couple (Dr+1,Er+1, αr+1, βr+1, γr+1) to be the derived couple of (Dr,Er, αr, βr, γr). For consistency of notation, we also call (D, E, α, β, γ) the first derived couple. Now we specialize the above proposition to the exact couple arising from a filtration to obtain the following corollary. Corollary 3.11. Let (D, E, α, β, γ) be the exact couple arising from a filtration (F p) of a complex (C, d), so α, β, γ have bidegrees (1, −1), (0, 0), (−1, 0) respectively. Then the rth derived couple (Dr,Er, αr, βr, γr) has the following properties: (i) the bigraded maps αr, βr, γr have bidegrees (1, −1), (1 − r, r − 1), (−1, 0), respectively; (ii) the differential dr has bidegree (−r, r − 1), and it is induced by βα−r+1γ; r+1 r r (iii) Ep,q = ker fp,q/im dp+r,q−r+1; r (iv) Dp,q = im (αp−1,q+1)(αp−2,q+2) ··· (αp−r+1,q+r−1); in particular, we have r p−1 p−2 p−r+1 p−r+1 p Dp,q = im (j j ··· j )∗ : Hn(F ) → Hn(F ) , where jp denotes the inclusion F p ,→ F p+1. INTRODUCTION TO SPECTRAL SEQUENCES 7

Proof. When we go to the next level, the bidegrees of α and γ stay the same, whereas the bidegree of β decreases by (1, −1). Therefore the bidegree of αr is still (−1, 1), the bidegree of γr is still (−1, 0), whereas the bidegree of βr is (1 − r, r − 1). For the diﬀerential dr, note that dr = βr−1γr−1 = βr−2(αr−2)−1γr−2 = βr−3(αr−3)−1α−1γ = ··· = βα−r+1γ. Hence the degree of dr is (−r + 1, r − 1) + (−1, 0) = (−r, r − 1). It follows that r r+1 ker dp,q Ep,q = r . im dp+r,q−r+1 2 Now for the last statement, recall that Dp,q = αp−1,q+1Dp−1,q+1. Thus 3 2 Dp,q = αp−1,q+1Dp−1,q+1 = αp−1,q+1αp−2,q+2Dp−2,q+2. It follows that r Dp,q = im αp−1,q+1αp−2,q+2 ··· αp−r+1,q+r−1. p p Now from the exact sequence in Proposition 3.9, we know that Dp,q = Hp+q(F ) and αp,q : Hp+q(F ) → p+1 p p p+1 Hp+q(F ) is just the induced map from the inclusion j : F → F . Therefore, r p−1 p−2 p−r+1 Dp,q = im (j j ··· j )∗, as claimed.

4. Spectral Sequences r r Definition 4.1. A spectral sequence is a sequence (E , d )r≥1 of differential bigraded modules such that Er+1 = H(Er, dr) for all r. We have proved the following theorem. Theorem 4.2. Every filtration of a complex produces a spectral sequence.

Proof. Combine Proposition 3.9 and Proposition 3.10 Definition 4.3. A filtration (F pC) of a complex (C, d) is bounded if for each n, there exist integers s = s(n) and t = t(n) such that s t (F C)n = {0} and (F C)n = Cn. Lemma 4.4. Let (Er, dr) be the spectral sequence arising from a bounded filtration of a complex. s r Then for each (p, q) ∈ Z × Z there exists an integer r (depending on p, q) such that Ep,q = Ep,q for all s > r. p p−1 Proof. For a fixed position (p, q), let n = p+q. Recall that Ep,q = Hn(F /F ) (see Proposition p p−1 3.9). For p > t(n), Fn /Fn = 0. Therefore p p−1 p p−1 Ep,q = Hn(F /F ) ⊆ Fn /Fn = 0. r r r−1 So Ep,q = 0. It follows that Ep,q = 0 for all r ≥ 1 and p > t(n) since Ep,q = Hp,q(E ). If p < s(n), p p p−1 r then Fn = 0 and so Fn /Fn = 0. Thus Ep,q = 0 for all r ≥ 1 and p < s(n). Now choose r big enough so that p+r > t(n+1) and p−r < s(n−1). Recall that the differential dr has bidegree (−r, r − 1). Therefore r r r+1 ker dp,q Ep,q r Ep,q = r = = Ep,q im dp+r,q−r+1 {0} r r because Ep−r,q+r−1 = 0 and Ep+r,q−r+1 = 0. INTRODUCTION TO SPECTRAL SEQUENCES 8

Definition 4.5. For the spectral sequence arising from a bounded filtration of a complex, we define ∞ r Ep,q : = Ep,q, where p, q, r are as in the above lemma. p p p p p Given a filtration (F C)p∈Z of a complex (C, d) with inclusions in :(F C)n → Cn, then i∗ : H(F ) → p p+1 p p+1 p H(C). Since F ⊆ F , we have im i∗ ⊆ im i∗ ; that is, (im i∗) is a filtration of H(C). p p p Definition 4.6. Let (F C)p∈Z is a filtration of a complex (C, d) and i : F → C are inclusions. Define p p p p Φ Hn(C) = im i∗, i∗ : Hn(F ) → Hn(C). p We call (Φ Hn(C))p∈Z the induced filtration of Hn(C). p Lemma 4.7. If (F C)p∈Z is a bounded filtration of a complex (C, d), then the induced filtration p (Φ H(C))p∈Z on homology is also bounded with the same bound. More precisely, if for n ∈ Z there exist s, t depending on n such that s s+1 t {0} = F Cn ⊆ F Cn ⊆ · · · ⊆ F Cn = Cn, then s s+1 t {0} = Φ Hn(C) ⊆ Φ Hn(C) ⊆ · · · ⊆ Φ Hn(C) = Hn(C). s s s t t t Proof. Since Fn = 0, Hn(F ) = 0 and so im i∗ = 0. Since Fn = Cn, i : F Cn → Cn is just t t t t the identity map. Thus i∗ : Hn(F ) → Hn(C) is also the identity map. Therefore i∗(Hn(F )) = Hn(C). r r Definition 4.8. A spectral sequence (E , d )r≥1 converges to a graded module H, denoted by 2 E ⇒ Hn, p,q p p if there is some bounded filtration (Φ H)p∈Z of H such that ∞ ∼ p p−1 Ep,q = Φ Hn/Φ Hn for all n, where we use the convention n = p + q. p Theorem 4.9. Let (F C)p∈Z be a bounded filtration of a complex (C, d) (so that the induced p r r filtration (Φ H) is bounded with the same bound) and (E , d )r≥1 be the corresponding spectral sequence. Then 2 E ⇒ Hn(C). p,q p 2 Proof. Fix a position (p, q) ∈ Z , let n = p + q. Consider the exact sequence obtained from the rth derived couple: r r r αr r β r γ r Dp+r−2,# −→ Dp+r−1,# −→ Ep,q −→ Dp−1,q. Recall that (see Corollary 3.11) r p−1 p−2 p−r+1 p−r+1 p Dp,q = im (j j ··· j )∗ : Hn(F ) → Hn(F ) , where jp denotes the inclusion F p ,→ F p+1. Replacing p by p + r − 1 and p + r − 2 we obtain r p+r−2 p+r−3 p p p+r−1 Dp+r−1,# = im (j j ··· j )∗ : Hn(F ) → Hn(F ) , and r p+r−3 p+r−4 p−1 p−1 p+r−2 Dp+r−2,# = im (j j ··· j )∗ : Hn(F ) → Hn(F ) . p+r−1 p+r−2 For large r such that p + r − 2 > t(n), we have Fn = Fn = Cn and the composite p+r−2 p p p r p p j ··· j of inclusions is just the inclusion i : Fn → Cn. Therefore, Dp+r−1,# = im i∗ = Φ Hn. r p−1 Similarly, Dp+r−2,# = Φ Hn. Thus we can rewrite the above exact sequence as p−1 p r r Φ Hn −→ Φ Hn −→ Ep,q −→ Dp−1,q. INTRODUCTION TO SPECTRAL SEQUENCES 9

Now r p−2 p−3 p−r p−r p−1 Dp−1,q = im (j j ··· j )∗ : Hn(F ) → Hn(F ) . p−r r But we can choose r big enough so that p − r < s(n) to make Fn = 0. For such r, Dp−1,q = 0 and so by the first isomorphism theorem r p p−1 Ep,q = Φ Hn/Φ Hn, which proves the theorem. 5. Spectral Sequences of a Total Complex Now we specialize our discussion above to the first and second filtrations of the total complex of a special type of bicomplex. 0 00 Definition 5.1. A bicomplex (M, d , d ) is called a first quadrant bicomplex if Mp,q = 0 whenever p < 0 or q < 0. Lemma 5.2. If (Tot(M),D) is the total complex of a first quadrant complex (M, d, d00), then the first and second filtrations of Tot(M) are bounded. I n Proof. From the definition of the first filtration of a total complex, we have F Tot(M)n = Tot(M)n I −1 II n and F Tot(M)n = {0}. Similarly, for the second filtration, we have F Tot(M)n = Tot(M)n II −1 and F Tot(M)n = {0} (look at the corresponding lines in the lattice). Theorem 5.3. Let M be a first quadrant bicomplex and let IEr and IIEr be the spectral sequences arising from the first and second filtrations of Tot(M), respectively. Then I 2 II 2 E ⇒ Hn(Tot(M)) and E ⇒ Hn(Tot(M)). p,q p p,q p Proof. This is a direct consequence of Theorem 4.9 and Lemma 5.2. 0 00 00 Let (M, d , d ) be a bicomplex. Recall that each pth column (Mp,∗, d ) forms a complex, therefore we can take homology of pth columns to obtain a bigraded module, which we denote by H00(M), 00 00 whose (p, q) term is Hq(Mp,∗). For each fixed q, the qth row H (M)∗,q of H (M)

...,Hq(Mp−1,∗),Hq(Mp,∗),Hq(Mp+1,∗),... 0 can be made into a complex if we define d p,q : Hq(Mp,∗) → Hq(Mp−1,∗) by 0 0 d p,q([z]) = [dp,q(z)], for [z] ∈ Hq(Mp,∗). 0 (Check that d p,q is well-defined, using the fact that each square of the bicomplex anticommutes.) Now we can take homology of qth rows of H00(M) to obtain a bigrade module, which we denote by 0 00 00 0 00 H H (M), whose (p, q) term is Hp(H (M)∗,q). We call the bigraded module H H (M) the first 0 00 iterated homology of M and denote its (p, q) term by HpHq (M). Proposition 5.4. If (M, d0, d00) is a first quadrant bicomplex, then I 1 I 2 0 00 E = Hq(Mp,∗) and E = H H (M) ⇒ Hn(Tot(M)). p,q p,q p q p Proof. Recall that (IEr) arises from the first filtration (IF pTot(M)) of the total complex (Tot(M),D) of M. To simplify notation, we omit the prescript I for the rest of this proof and abbreviate I p p F Tot(M)n to Fn . Recall also that (see Proposition 3.9) 1 p p−1 Ep,q = Hn(F /F ). p p−1 p p−1 p p Now the map Dn : Fn /Fn → Fn−1/Fn−1 is induced from Dn : Fn → Fn−1, where Dn : Tot(M)n → Tot(M)n−1 given by X 0 00 Dn = (dp,q + dp,q). p+q=n INTRODUCTION TO SPECTRAL SEQUENCES 10

p p−1 p p−1 By definition of the first filtration, Fn /Fn = Mp,q and Fn−1/Fn−1 = Mp,q−1 (here again we use p p−1 the convention that n = p + q). To investigate the map Dn, pick an element an ∈ Fn /Fn = Mp,q. p Then think of an as an element in Fn , we have 0 00 00 p−1 Dnan = Dnan = dp,qan + dp,qan ≡ dp,qan mod Fn−1 0 p−1 00 since dp,qan ∈ Fn−1 . Thus Dn = dp,q and so 00 1 p p−1 ker Dn ker dp,q Ep,q = Hn(F /F ) = = 00 = Hq(Mp,∗). im Dn+1 im dp,q+1 2 1 1 1 Now to compute Ep,q, recall (Proposition 3.10) that the differential d : Ep,q → Ep−1,q is given by 1 γ β 1 Ep,q −→ Dp−1,q −→ Ep−1,q, 1 p p−1 p−1 p−1 p−2 where Ep,q = Hn(F /F ), Dp−1,q = Hn−1(F ), Ep−1,q = Hn−1(F /F ), γ is the connecting homomorphism and β is induced from the quotient map F p−1 → F p−1/F p−2. Pick 00 [z] ∈ Hq(Mp,∗), where z ∈ Mp,q such that dp,qz = 0. The connecting homomorphism arises from the following diagram,

Mp,q Mp,q 0

0 Mp−1,q Mp−1,q

0 1 1 1 which is easily seen to be just dp,q. Therefore the map d : Ep,q = Hq(Mp,∗) → Ep−1,q = Hq(Mp−1,∗) is given by 1 0 d ([z]) = [dp,qz] for [z] ∈ Hq(Mp,∗). Thus ker d0 E2 = p,q = H0 H00(M), p,q 0 p q im d p+1,q as required. Now we describe an analogous construction. Let (M, d0, d00) be a bicomplex. Recall that each pth 0 row (M∗,p, d ) forms a complex, therefore we can take homology of pth rows to obtain a bigraded 0 module, which we denote by H (M), whose (p, q) term is Hq(M∗,p). For each ﬁxed q, the qth 0 0 column H (M)q,∗ of H (M)

...,Hq(M∗,p−1),Hq(M∗,p),Hq(M∗,p+1),... 00 can be made into a complex if we define d p,q : Hq(M∗,p) → Hq(M∗,p−1) by 00 00 d p,q([z]) = [dp,q(z)], for [z] ∈ Hq(M∗,p). 00 (Check that d p,q is well-defined, using the fact that each square of the bicomplex anticommutes.) Now we can take homology of qth columns of H0(M) to obtain a bigrade module, which we denote 00 0 0 00 0 by H H (M), whose (p, q) term is Hp(H (M)q,∗). We call the bigraded module H H (M) the 00 0 second iterated homology of M and denote its (p, q) term by Hp Hq(M). Proposition 5.5. If (M, d0, d00) is a first quadrant bicomplex, then II 1 II 2 00 0 E = Hq(M∗,p) and E = H H (M) ⇒ Hn(Tot(M)). p,q p,q p q p INTRODUCTION TO SPECTRAL SEQUENCES 11

Proof. Since the second filtration of Tot(M) is the first filtration of Tot(M t), (IIEr) is the same as (IEer), the spectral sequence arising from the first filtration of Tot(M t) (note that M t is still a first quadrant bicomplex). From Proposition 5.4, we have

II 1 I 1 t Ep,q = Eep,q = Hq(Mp,∗) = Hq(M∗,p), and II 2 I 2 00 t 0 00 0 Ep,q = Eep,q = Hp(H (M )∗,q) = Hp(H (M)q,∗) = Hp Hq(M), t 0 00 0 00 00 where we remind the readers that for the transpose bicomplex (M , δ , δ ), δp,q = dq,p and δp,q = 0 dq,p.

r r 2 Deﬁnition 5.6. A spectral sequence (E , d ) collapses on the p-axis if Ep,q = {0} for all q 6= 0 (hence all the non-zero modules lie on the p-axis). A spectral sequence (Er, dr) collapses on the 2 q-axis if Ep,q = {0} for all p 6= 0 (hence all the non-zero modules lie on the q-axis).

Proposition 5.7. Let (Er, dr) be a ﬁrst quadrant spectral sequence (so it’s either IEr or IIEr). Then r r ∞ 2 (i) if (E , d ) collapses on either axis, then Ep,q = Ep,q for all p, q; r r ∼ 2 (ii) if (E , d ) collapses on the p-axis, then Hn(Tot(M)) = En,0; r r ∼ 2 if (E , d ) collapses on the q-axis, then Hn(Tot(M)) = E0,n. Proof. For part (i), suppose (Er, dr) collapses on the p axis. Consider a position (p, q). If q 6= 0, i.e. 2 ∞ 2 not on the p axis, then Ep,q = 0. It follows that Ep,q = Ep,q = 0. If q = 0, i.e. on the p axis, then notice that the map dr has bidegree (−r, r − 1) and so it’s never horizontal for r > 2. It follows that r r+1 ker dp,q r Ep,0 = r = Ep,0 for r ≥ 2 im dp+r,−r+1

r r ∞ 2 because the source of dp+r,−r+1 and the target of dp,q are both 0. Thus Ep,0 = Ep,0. A similar argument applies for the case where (Er, dr) collapses on the q-axis. Now for part (ii), let’s assume that (Er, dr) collapses on the q-axis. Since M is ﬁrst quadrant, we have the following ﬁltration of Hn(Tot(M)):

−1 0 n {0} = Φ Hn ⊆ Φ Hn ⊆ · · · ⊆ Φ Hn = Hn.

2 From Theorem 5.3, we know that E ⇒ Hn(Tot(M)) (where n = p + q). In other words, p,q p

p p−1 ∞ Φ Hn/Φ Hn = Ep,q for p + q = n.

r r ∞ From part (i), since (E , d ) collapses on the q-axis, we have Ep,q = 0 for p > 0. It follows that

0 1 n Φ Hn = Φ Hn = ··· = Φ Hn = Hn.

∞ 2 When p = 0, we have E0,q = E0,p. Therefore,

2 0 −1 0 E0,p = Φ Hn/Φ Hn = Φ Hn.

2 r r Thus E0,p = Hn(Tot(M)). The case where (E , d ) collapses on the p axis can be proved similarly. INTRODUCTION TO SPECTRAL SEQUENCES 12

6. An Application As an application of spectral sequences, we give a proof that Tor is independent of the variable resolved. We first recall the definition of Tor. Definition 6.1. An R-module P is projective if for any R-modules G and H and homomorphisms α: P → H and surjection β : G → H, there exists a homomorphism φ making the diagram below commute.

P φ α

G H 0 β

Lemma 6.2. Every free R-module is projective.

Proof. Suppose P is free. Let B = {bi : i ∈ I} be a basis for P . Since β is surjective, for each bi ∈ B, there exists gi ∈ G such that β(gi) = α(bi). Define φ: B → G by φ(bi) = gi (note that we have to invoke Axiom of Choice here) and extend by linearity. This gives a well-defined function φ: P → G (see [1, Proposition 2.34]). Now to check that φ makes the diagram commute, pick P a = i ribi ∈ P . Then ! ! X X X X β(a) = βφ ribi = β riφ(bi) = riβ(gi) = riα(bi) = α(a), i i i i as required. P Definition 6.3. Let X = {xi : i ∈ I} be a basis of a free R-module F and let R = { i rjixi : j ∈ J} be a subset of F . If K is the submodule of F generated by R, then we say that the module M = F/K has generators X and relations R and the ordered pair hX |Ri a presentation of M. Lemma 6.4. Every R-module M has a presentation. Proof. Let F be the free module generated by all the elements of M. Define φ: F → M by φ(m) = m for m ∈ M and extend by linearity. Clearly φ is surjective. By the first isomorphism theorem, we have M = F/ ker φ, i.e. M = hF | ker φi. Definition 6.5. Let A be an R-module, a projective resolution of A is an exact sequence

d2 d1 ε P = · · · −→ P2 −→ P1 −→ P0 −→ A −→ 0 in which each Pn is projective. A deleted projective resolution of A is the complex

d2 d1 PA = · · · −→ P2 −→ P1 −→ P0 −→ 0 Proposition 6.6. Every R-module A has a projective resolution. Proof. We show that A has a free resolution, which in turn implies that A has a projective resolution since each free module is projective. We know that A has a presentation hF0|K1i, which gives a short exact sequence: i1 0 −→ K1 −→ F0 −→ A −→ 0.

Similarly, K1 has a presentation hF1|K2i, which gives a short exact sequence:

i2 1 0 −→ K2 −→ F1 −→ K1 −→ 0. INTRODUCTION TO SPECTRAL SEQUENCES 13

Now we splice these two sequences together to obtain

i2 i11 0 −→ K2 −→ F1 −−→ F0 −→ A −→ 0.

We need to check exactness at F1 and F0. Note that since 1 is surjective,

i11(F1) = i1(K1) = ker .

Since i1 is injective, ker i11 = ker 1 = im 2. So the sequence is exact. This procedure can be continued indeﬁnitely and we obtain a free resolution of A.

Before we proceed, we briefly recall the definition of a left derived functor. Let ModR denote the category of R-modules (the definitions work for more general abelian categories, but here we just focus on the category of R-modules for simplicity). Let T : ModR → ModR be a covariant, right exact functor, i.e. for any short exact sequence of R-modules ψ φ 0 −→ L −→ M −→ N −→ 0, the following sequence T ψ T φ T (L) −→ T (M) −→ T φT (N) −→ 0 is also exact (notice 0 lies on the right). Now for an R-module A, choose a deleted projective resolution of A: d2 d1 PA = · · · −→ P2 −→ P1 −→ P0 −→ 0. Then apply the functor T to PA to obtain the complex TPA:

T (d2) T (d1) · · · −→ T (P2) −→ T (P1) −→ T (P0) −→ 0. Now for each R-module A, the functors

(LnT )A: = Hn(TPA) are called the left derived functors of T (the action of LnT on morphisms is a bit more compli- cated and we do not need it in this notes). One important property of left derived functors is given in the following proposition.

Proposition 6.7. Let A be an R-module and suppose PeA is another deleted projective resolution of A and let LenT be the corresponding left derived functors. Then ∼ (LnT )A = (LenT )A for all n ≥ 0. In other words, the deﬁnition of left derived functors does not depend on the choice of a deleted projective resolution of A.

Proof. For a proof, we refer the readers to [1, Proposition 6.20]. Now we specialize our discussion to the functors T = ⊗R B and S = A ⊗R . Proposition 6.8. The functors R and S are right exact.

Proof. We refer the readers to [1, Theorem 2.63] for a proof.

Deﬁnition 6.9. If B is a left R-module and T = ⊗R B, deﬁne R Torn (,B) = LnT. In concrete terms, for each right R-module A, choose a deleted projective resolution of A:

d2 d1 PA = · · · −→ P2 −→ P1 −→ P0 −→ 0.

Now form the complex PA ⊗R B:

d2⊗1 d1⊗1 · · · −→ P2 ⊗R B −−−→ P1 ⊗R B −−−→ P0 ⊗R B −→ 0. INTRODUCTION TO SPECTRAL SEQUENCES 14

Then R Torn (A, B) = Hn(PA ⊗R B).

Deﬁnition 6.10. If A is a right R-module and S = A ⊗R , deﬁne R torn (A, ) = LnS. In concrete terms, for each left R-module B, choose a deleted projective resolution of B:

d2 d1 QB = · · · −→ Q2 −→ Q1 −→ Q0 −→ 0.

Now form the complex A ⊗R QB:

1⊗d2 1⊗d1 · · · −→ A ⊗R Q2 −−−→ A ⊗R Q1 −−−→ A ⊗R Q0 −→ 0. Then R torn (A, B) = Hn(A ⊗R QB).

Definition 6.11. A right R-module A is flat if A ⊗R is an exact functor; that is, whenever ψ φ 0 −→ L −→ M −→ N −→ 0 is a short exact sequence of R-modules, the sequence 1⊗ψ 1⊗φ 0 −→ A ⊗R L −→ A ⊗R M −→ A ⊗R N −→ 0 is also exact. A left R-module B is flat if ⊗R B is an exact functor. Proposition 6.12. Every projective R-module is flat.

Proof. We refer the readers to [1, Proposition 3.46]. Remark 6.13. Let A be a ﬂat module, then it means in particular that if ψ φ L −→ M −→ N is exact at M, then 1⊗ψ 1⊗φ A ⊗R L −→ A ⊗R M −→ A ⊗R N is exact at A ⊗R M. This is because we have the short exact sequence 0 −→ im ψ −→ M −→ im φ −→ 0. Since A is ﬂat we obtain the short exact sequence

0 −→ A ⊗R im ψ −→ A ⊗R M −→ A ⊗R im φ −→ 0. Therefore ker(1 ⊗ φ) ker(1 ⊗ φ) = = 0. im (1 ⊗ ψ) A ⊗R im ψ A similar argument works for the case ⊗R B. Given two complexes 0 ∆p PA = · · · −→ Pp −→ Pp−1 −→ · · · −→ P0 −→ 0 and 00 ∆q QB = · · · −→ Qq −→ Qq−1 −→ · · · −→ Q0 −→ 0, we can form a ﬁrst quadrant bicomplex (M, d0, d00) whose terms are given by 0 0 00 p 00 Mp,q = Pp ⊗R Qq, dp,q = ∆p ⊗ 1, dp,q = (−1) 1 ⊗ ∆q . The total complex of M is called the tensor product of complexes and is denoted by Tot(M) = PA ⊗R QB. INTRODUCTION TO SPECTRAL SEQUENCES 15 ......

0 P0 ⊗ Q2 P1 ⊗ Q2 P2 ⊗ Q2 ···

0 P0 ⊗ Q1 P1 ⊗ Q1 P2 ⊗ Q1 ···

0 P0 ⊗ Q0 P1 ⊗ Q0 P2 ⊗ Q0 ···

0 0 0

0 00 00 0 To see that M is indeed a bicomplex, we just need to check that dp,q−1dp,q + dp−1,qdp,q = 0. Plug in the deﬁnitions of d0 and d00 we obtain 0 p 00 p−1 00 0 (∆p ⊗ 1)(−1) (1 ⊗ ∆q ) + (−1) (1 ⊗ ∆q )(∆p ⊗ 1), which is 0, as can easily be checked using the deﬁnition of tensor product.

Theorem 6.14. Let PA and QB be deleted projective resolutions of a right R-module A and a left R ∼ R R-module B, then Torn (A, B) = torn (A, B). 0 00 Proof. Let (M, d , d ) be the bicomplex as in the above paragraph whose total complex is PA ⊗ QB. The main idea of the proof is to look at IEr and IIEr. Let us ﬁrst compute IEr using the ﬁrst iterated homology. The pth column of M is

Mp,∗ =−→ Pp ⊗ Qq+1 −→ Pp ⊗ Qq −→ Pp ⊗ Qq−1 −→ .

Since Pp is projective, hence ﬂat, this sequence is exact at every q > 0 (see Remark 6.13). Therefore Hq(Mp,∗) = 0 for q > 0 and

H0(Mp,∗) = coker(Pp ⊗R Q1 → Pp ⊗R Q0).

Now Q1 −→ Q0 −→ B −→ 0 is exact, so by right exactness of tensor the sequence

Pp ⊗R Q1 −→ Pp ⊗R Q0 −→ Pp ⊗R B −→ 0 is also exact. Therefore

H0(Mp,∗) = coker(Pp ⊗R Q1 → Pp ⊗R Q0) = Pp ⊗R B by the ﬁrst isomorphism theorem. To sum up, ( I 1 0 if q > 0, Ep,q = Hq(Mp,∗) = Pp ⊗R B if q = 0. Therefore, ( I 2 0 00 0 if q > 0, Ep,q = HpHq (M) = R Hp(PA ⊗ B) = Torp (A, B) if q = 0. So the spectral sequence (IEr) collapses on the p-axis. It follows from Proposition 5.7 that I 2 R Hn(Tot(M)) = En,0 = Torn (A, B). Now we compute IIEr using the second iterated homology. The pth row of M is

M∗,p =−→ Pq+1 ⊗ Qp −→ Pq ⊗ Qp −→ Pq−1 ⊗ Qp −→ .

Since Qp is projective, hence ﬂat, we have Hq(M∗,p) = 0 for q > 0 and

H0(M∗,p) = coker(P1 ⊗R Qp → P0 ⊗R Qp). INTRODUCTION TO SPECTRAL SEQUENCES 16

By the same argument as above we obtain

H0(M∗,p) = A ⊗R QB. To sum up, ( II 1 0 if q > 0, Ep,q = Hq(M∗,p) = A ⊗R Qp if q = 0. Therefore, ( II 2 00 0 0 if p > 0, Ep,q = Hp Hq(M) = R Hq(A ⊗ QB) = torq (A, B) if p = 0. So the spectral sequence (IIEr) collapses on the q-axis. It follows from Proposition 5.7 that II 2 R Hn(Tot(M)) = E0,n = torn (A, B). Thus we conclude that R R Torn (A, B) = Hn(Tot(M)) = torn (A, B), and so it does not matter which variable we resolve ﬁrst. References [1] Joseph J. Rotman. An Introduction to Homological Algebra. Springer, second edition, 2009. [2] Joseph J. Rotman. Advanced Modern Algebra. American Mathematical Society, second edition, 2010.